The Borsuk-Ulam property for homotopy classes of maps between the torus and the Klein bottle

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The Borsuk-Ulam property for homotopy classes of maps between the torus and the Klein bottle

Daciberg Lima Gonçalves, John Guaschi, Vinicius Casteluber Laass

To cite this version:

Daciberg Lima Gonçalves, John Guaschi, Vinicius Casteluber Laass. The Borsuk-Ulam property for homotopy classes of maps between the torus and the Klein bottle. Topological Methods in Nonlin- ear Analysis, Juliusz Schauder University Centre for Nonlinear Studies, 2020, 56 (2), pp.529–558.

�10.12775/TMNA.2020.003�. �hal-02404504�

The Borsuk-Ulam property for homotopy classes of maps between the torus and the Klein bottle

DACIBERG LIMA GONÇALVES

JOHN GUASCHI

VINICIUS CASTELUBER LAASS

17th November 2019

Abstract

LetM be a topological space that admits a free involutionτ, and let N be a topological space. A homotopy class β ∈[M, N]is said to havethe Borsuk-Ulam property with respect to τ if for every representative map f : M → N of β, there exists a point x ∈ M such that f(τ(x)) =f(x). In this paper, we determine the homotopy classes of maps from the 2-torus T² to the Klein bottleK² that possess the Borsuk-Ulam property with respect to a free involution τ1 of T² for which the orbit space is T². Our results are given in terms of a certain family of homomorphisms involving the fundamental groups of T² and K².

1 Introduction

In the early twentieth century, St. Ulam conjectured that if f : Sⁿ → Rⁿ is a continuous map, there exists x ∈ Sⁿ such that f(A(x)) = f(x), where A : Sⁿ → Sⁿ is the antipodal map. The confirmation of this result by K. Borsuk in 1933 [1], known as the Borsuk-Ulam theorem, was the beginning of what it now referred to as Borsuk-Ulam type theorems or theBorsuk-Ulam property.

More information about the history and some applications of the Borsuk-Ulam theorem may be found in [9], for example.

One possible generalisation of the classical Borsuk-Ulam theorem is to substitute Sⁿ and Rⁿ by other spaces, and to replace the antipodal map by a free involution. A natural question is the following: does every continuous map collapse an orbit of the involution? More precisely, given topological spaces M and N such that M admits a free involution τ, we say that the triple (M, τ;N) has the Borsuk-Ulam property if for every continuous map f : M → N, there exists a point x ∈ M such that f(τ(x)) = f(x). For the cases where M is a compact surface without boundary admitting a free involutionτ andN is eitherR² or a compact surface without boundary, the triples (M, τ;N) that have the Borsuk-Ulam property have been classified (see [3] and [4]).

One generalisation of this property is to consider a local Borsuk-Ulam problem in the sense of the following definition: a homotopy class β ∈[M, N]has the Borsuk-Ulam property with respect to τ if for every representative f :M →N of β, there exists a point x∈M such thatf(τ(x)) =f(x).

∗Departamento de Matemática, IME, Universidade de São Paulo, Caixa Postal 66281, Ag. Cidade de São Paulo, CEP: 05314-970, São Paulo, SP, Brazil. e-mail: dlgoncal@ime.usp.br

†Normandie Univ., UNICAEN, CNRS, Laboratoire de Mathématiques Nicolas Oresme UMR CNRS 6139, 14000 Caen, France. e-mail: john.guaschi@unicaen.fr

‡Departamento de Matemática, IME, Universidade Federal da Bahia, Av. Adhemar de Barros, S/N Ondina CEP: 40170-110, Salvador, BA, Brazil. e-mail: vinicius.laass@ufba.br

In [5], the Borsuk-Ulam problem for homotopy classes of maps between compact surfaces without boundary was studied, and the sets [T²,T²] and [K²,K²] whose elements possess the Borsuk-Ulam property were characterised. By [4, Theorem 12], for any involution τ : T² → T², the triple(T², τ;K²)does not have the Borsuk-Ulam property. Using this information, in this paper we classify the homotopy classes of maps from T² to K² that have the Borsuk-Ulam property for the orientation-preserving free involutionτ₁ ofT². Our approach, which we now describe, is similar to that used in [5]. First, as in [5, Theorems 12 and 19], we identify π₁(T²,∗) and π₁(K²,∗)with the free Abelian group Z⊕Zand the (non-trivial) semi-direct product Z ⋊ Z respectively. These identifications will be helpful in formulating the results and in making explicit computations.

To prove our results, we will make use of the following algebraic description given in [6, Corollary 2.1] of the set [T²,K²] in terms of pointed homotopy classes and the corresponding fundamental groups.

Proposition 1.1. The set [T²,K²] is in bijection with the subset of Hom(Z⊕Z,Z ⋊ Z) whose elements are described as follows:

Type 1:

(

(1,0)7→(i,2s₁+ 1) (0,1)7→(0,2s₂) Type 2:

(

(1,0)7→(i,2s1+ 1) (0,1)7→(i,2s2+ 1)

Type 3:

(

(1,0)7→(0,2s₁) (0,1)7→(i,2s₂+ 1) Type 4:

(

(1,0)7→(r1,2s1) (0,1)7→(r2,2s2),

where i∈ {0,1} and s₁, s₂ ∈Z for Types 1,2 and 3, and r₁, r₂, s₁, s₂ ∈Z and r₁ ≥0 for Type 4.

Remark 1.2. The bijection of Proposition 1.1 may be obtained using standard arguments in homotopy theory that are described in detail in [11, Chapter V, Corollary 4.4], and more briefly in [5, Theorem 4]. In our specific case, the bijection is defined as follows: given a homotopy class β ∈[T²,K²], there exists a pointed map f: (T²,∗)→(K²,∗) that gives rise to a representative of β if we omit the basepoints. The induced homomorphism f_#: π₁(T²,∗)→π₁(K²,∗)is conjugate to exactly one of the elements ofHom(Z⊕Z,Z⋊Z)described in Proposition1.1, which we denote by β_#. Note that β_# does not depend on the choice of f.

In order to solve the Borsuk-Ulam problem for homotopy classes, we now describe the relevant involution of T². Consider the following short exact sequence:

1→π₁(T²) =Z⊕Z−→ⁱ¹ π₁(T²) =Z⊕Z−→^θ1 Z2 →1 (1) where:

i₁ : (

(1,0)7−→(2,0) (0,1)7−→(0,1) θ₁ :

(

(1,0)7−→1 (0,1)7−→0.

By standard results in covering space theory, there exists a double covering c₁ : T² → T² whose induced homomorphism on the level of fundamental groups is i₁. If τ₁ : T² → T² is the non- trivial deck transformation associated with c1, then τ1 is a free involution. Further, τ1 lifts to a homeomorphism bτ₁ : R² → R², where τb₁(x, y) = (x+ ¹₂, y) for all (x, y) ∈ R². In this paper, we classify the elements of the set[T²,K²]that possess the Borsuk-Ulam property with respect to τ₁. This is achieved in the following theorem, which is the main result of this paper.

Theorem 1.3. Given a non-zero integer t, let e(t) be its 2-adic evaluation. With the notation of Proposition 1.1, let β ∈ [T²,K²] and β_# ∈Hom(Z⊕Z,Z ⋊ Z). Then β has the Borsuk-Ulam property with respect to τ1 if and only if one of the following conditions is satisfied:

(a) β_# is a homomorphism of Type 3.

(b) β_# is a homomorphism of Type 4, where s₁ is odd and r₂ 6= 0, and additionally e(r₁)> e(r₂) if r₁ 6= 0.

Besides the introduction and an Appendix, this paper consists of three sections. In Section 2, we show how to reduce the number of homotopy classes to be studied with respect to the Borsuk- Ulam property. In Section 3, we study the normal closure of σ² in P₂(K²), which is a free group of infinite rank. A convenient basis for this subgroup is obtained in the Appendix. In Section 4, we prove Theorem 1.3.

The study of the free involution τ₂ ofT² for which the associated orbit space is the Klein bottle is the subject of work in progress.

2 Some preliminary results

The following results will enable us to reduce the number of cases to be analysed in the proof of Theorems 1.3.

Lemma 2.1. Let M and N be topological spaces, let τ: M → M be a free involution, and let H: N →N be a homeomorphism. Then the mapH: [M, N]→[M, N]defined by H([f]) = [H◦f] for all maps f: M →N is a bijection. Further, if β ∈[M, N]is a homotopy class, thenβ has the Borsuk-Ulam property with respect to τ if and only if H(β) has the Borsuk-Ulam property with respect to τ.

Proof. Clearly the map H is a bijection whose inverse is given by H⁻¹([g]) = [H⁻¹◦g]. To prove the second part of the statement, let β ∈ [M, N] be a homotopy class that has the Borsuk-Ulam property with respect to τ, and let g ∈ H(β). Thus H⁻¹ ◦g ∈ β, and hence there exists x ∈ M such that H⁻¹◦g(x) = H⁻¹◦g(τ(x)). Therefore g(x) =g(τ(x)), and we conclude that H(β)has the Borsuk-Ulam property. The converse follows in a similar manner using H⁻¹.

Proposition 2.2. Let τ :T² → T² be a free involution, and let β, β^′ ∈ [T²;K²] such that β_#, β_#^′ are both of Type 1, 2 or 3. Suppose that the second coordinates of β_#(ω) and β_#^′ (ω) are equal for all ω ∈π₁(T²,∗) and the integer i that defines the homomorphism β_# (resp. β_#^′ ) is equal to 0 (resp. 1). Then β has the Borsuk-Ulam property with respect to τ if and only if β^′ does.

Proof. Let h : Z ⋊ Z → Z ⋊ Z be the homomorphism defined on the generators of Z ⋊ Z by h(1,0) = (1,0) and h(0,1) = (1,1). Then h is well defined, and it is an isomorphism whose inverse h⁻¹ : Z ⋊ Z → Z ⋊ Z is given by h⁻¹(1,0) = (1,0) and h⁻¹(0,1) = (−1,1). By [12, Theorem 5.6.2], there exists a homeomorphism H : (K²,∗) → (K²,∗) such that H_# = h and H_#⁻¹ = h⁻¹. Suppose that β_# and β_#^′ are both of Type 1, and that they satisfy the hypothesis of the statement, and let f: (T²,∗) → (K²,∗) be a representative map of β. Without loss of generality, we may suppose that f_# = β_#. Assume that β has the Borsuk-Ulam property with respect to τ. Then:

(H◦f)_#(1,0) = h(β_#(1,0)) =h(0,2s₁+ 1) =h(0,1)^2s1+1

= ((1,1)(1,1))^s1(1,1) = (0,2)^s1(1,1) = (1,2s₁+ 1),and

(H◦f)#(0,1) = h(β#(0,1)) =h(0,2s2) =h(0,1)^2s2 = ((1,1)(1,1))^s2 = (0,2)^s2 = (0,2s2).

Then H◦f is a representative of β^′, and the conclusion follows from Lemma 2.1. The converse follows in a similar way using H⁻¹ insteadH. The arguments for the cases of homomorphisms of Types 2 and 3 are analogous, and the details are left to the reader.

Remark 2.3. Using Lemma 2.1, one may show that Proposition 2.2 holds in more generality.

However the above statement will be sufficient for our purposes.

3 The normal closure of σ

in P

( K

)

Let S be a compact, connected surface without boundary. The ordered 2-point configuration space of S is defined by F2(S) = {(x, y)∈ S×S | x 6=y}, D2(S) is the orbit space of F2(S) by the free involution τ_S : F₂(S) → F₂(S) defined by τ_S(x, y) = (y, x), and P₂(S) = π₁(F₂(S)) and B₂(S) = π₁(D₂(S)) are the pure and full 2-string braid groups of S respectively [2]. We have a short exact sequence:

1→P₂(S)→B₂(S)→^π Z2 →1, (2)

where π :B2(S)→Z2 is the homomorphism that to an element of B2(S) associates its permuta- tion. If p₁ :F₂(S)→S is the projection onto the first coordinate, the map(p₁)_#:P₂(S)→π₁(S) may be interpreted geometrically as the surjective homomorphism that forgets the second string.

Letσ ∈B2(S)\P2(S)be the Artin generator ofB2(S)that geometrically swaps the two basepoints.

Then σ² ∈ P₂(S), and the normal closure of σ² in P₂(S), which we denote by hσ²i, is also the normal closure of σ² in B₂(S) by (2). In this section, we shall take S to be the Klein bottle, and we will show that hσ²i is a free group of countably-infinite rank for which we shall exhibit a basis.

We will also express certain elements of hσ²i in this basis.

The following proposition summarises some results of [5, Section 4] regarding the structure of P₂(K²)and the action by conjugation by σ on this group.

Proposition 3.1. [5, Theorems 19 and 20] The group P₂(K²) is isomorphic to the semi-direct product F(u, v)⋊θ (Z ⋊ Z), where F(u, v) is the free group of rank 2 on the set {u, v} and the action θ:Z ⋊ Z→Aut(F(u, v)) is defined as follows:

θ(m, n) :







u7→B^m−δnu^εnB^−m+δn v 7→B^mvu^−2mB^−m+δn B 7→B^εn,

where δ_n = (

0 if n is even

1 if n is odd, ε_n = (−1)ⁿ and B = uvuv⁻¹. With respect to this description, the following properties hold:

• the element σ ∈B₂(K²) satisfies σ² = (B; 0,0).

• if l_σ :P₂(K²)→ P₂(K²) is the homomorphism defined by l_σ(b) = σbσ⁻¹ for all b ∈P₂(K²), then:

l_σ(u^r; 0,0) = ((Bu⁻¹)^rB^−r;r,0) l_σ(1;m,0) = (1;m,0) lσ(v^s; 0,0) = ((uv)^−s(uB)^δs; 0, s) lσ(1; 0, n) = (B^δn; 0, n) l_σ(B; 0,0) = (B; 0,0)

for all m, n, r, s∈Z, where the symbol 1 denotes the trivial element of F(u, v).

• the homomorphism (p₁)_#:P₂(K²)→π₁(K²) = Z ⋊ Z satisfies (p₁)_#(w;r, s) = (r, s).

From now on, we identify P2(K²) with F(u, v)⋊θ(Z ⋊ Z) without further comment.

Remark 3.2. It follows from Proposition 3.1 that for all m, n ∈Z, the automorphism θ(m, n) : F(u, v)→F(u, v) depends only on m and the parity of n, in particular θ(m, n) = θ(m, δ_n).

Consider the following maps:

ι:

(F(u, v)−→P₂(K²)

w7−→(w; 0,0) and p_F :

( P₂(K²)−→F(u, v) (w;m, n)7−→w.

Note thatιis a homomorphism, but due to the actionθ,p_F is not. Consider the mapρ:F(u, v)→ F(u, v) defined by:

ρ=pF ◦lσ ◦ι, (3)

and the homomorphism g :F(u, v)→Z ⋊ Z defined on the basis {u, v} by:

(

g(u) = (1,0)

g(v) = (0,1). (4)

Using Proposition 3.1 and (3), we obtain the following commutative diagram:

F(u, v)

F(u, v) ^{ι //}

gYYYYYYYYYYYYYYYY,, YY

YY YY YY YY YY YY Y

ρ

22e

ee ee ee ee ee ee ee ee ee ee ee ee ee ee ee ee ee

e P2(K²) ^{lσ //}P2(K²) =F(u, v)⋊θ(Z ⋊ Z)

(p1)#

pF

OO

Z ⋊ Z=π₁(K²),

(5)

from which it follows that:

l_σ(w; 0,0) = (ρ(w);g(w)) (6)

for all w∈F(u, v). Further if w, z∈F(u, v) then:

ρ(wz) = (p_F ◦l_σ)(wz; 0,0) =p_F(l_σ(w; 0,0). l_σ(z; 0,0)) =p_F (ρ(w);g(w)).(ρ(z);g(z))

=p_F ρ(w)θ(g(w))(ρ(z));g(w)g(z)

=ρ(w)θ(g(w))(ρ(z)).

Thus the map ρ:F(u, v)→F(u, v) is not a homomorphism, but ifw∈Ker(g)then:

ρ(wz) = ρ(w)ρ(z). (7)

In Theorem A.3of the Appendix, we prove that:

Ker(g) =

B_k,l :=v^ku^lBu^−lv^−k, k, l ∈Z |−

. (8)

Let w ∈ F(u, v) and (m, n) ∈ Z ⋊ Z. Using Proposition 3.1, we see that B ∈ Ker(g), and then that:

θ(m, n)(wBw⁻¹) =θ(m, n)(w)B^εnθ(m, n)(w)⁻¹ ∈Ker(g). (9) Further,

l_σ(wBw⁻¹; 0,0) = l_σ(w; 0,0)l_σ(B; 0,0)l_σ(w; 0,0)^{−1 (}= (ρ(w);⁵⁾ g(w))(B; 0,0)(ρ(w);g(w))⁻¹

= (ρ(w)θ(g(w))(B);g(w))(θ(g(w)⁻¹)(ρ(w)⁻¹);g(w)⁻¹)

= (ρ(w)θ(g(w))(B)ρ(w)⁻¹; 0,0). (10)

Composing (10) by p_F, it follows from (5) and (9) that:

ρ(wBw⁻¹) =ρ(w)θ(g(w))(B)ρ(w)⁻¹ ∈Ker(g). (11) In particular, ρ(Bk,l) ∈ Ker(g) for all k, l ∈ Z, and using (7) and (8), the restriction of ρ to Ker(g) yields an endomorphism of Ker(g), which we also denote by ρ. Further, θ(m, n)(B_k,l) ∈ Ker(g) for all k, l, m, n ∈ Z using (9), and thus θ determines a homomorphism from Z ⋊ Z to Aut(Ker(g)), which we also denote by θ. Note that for all m, m ∈ Z, the endomorphism θ(m, n) : Ker(g) → Ker(g) is indeed surjective. To see this, let k, l ∈ Z, and let ξ ∈ F(u, v) be such that θ(m, n)(ξ) = v^ku^l. Then θ(m, n)(ξB^εnξ⁻¹) = B_k,l, and ξB^εnξ⁻¹ ∈ Ker(g) because B ∈ Ker(g). Hence the image of θ(m, n) : Ker(g) → Ker(g) contains the basis {Bk,l}k,l∈Z of Ker(g), and so this homomorphism is surjective.

The following result describes the subgroup hσ²i.

Proposition 3.3. The injective homomorphism ι: F(u, v) →P₂(K²) defined by ι(w) = (w; 0,0) for all w∈F(u, v), restricts to an isomorphism between Ker(g) andhσ²i, which we also denote by ι. In particular, hσ²i is a free group of infinite rank for which a basis is given by{(B_k,l; 0,0)}k,l∈Z, andl_σ(hσ²i)⊂Ker((p₁)_#). Up to this isomorphism, the homomorphismsθ :Z⋊Z→Aut(Ker(g)) and ρ:Ker(g)→Ker(g) induce homomorphisms Z ⋊ Z→Aut(hσ²i) and hσ²i → hσ²i, which we also denote by θ and ρ respectively. Further, the following diagram is commutative:

hσ²_{ _}i ^{ρ //}

hσ²_{ _}i

P₂(K²) ^{lσ //}P₂(K²).

Proof. LetH =ι(Ker(g))⊂P₂(K²). We start by showing that hσ²i =H. To see this, first note that σ² = (B; 0,0) and ι(B_k,l)∈ hσ²i for all k, l ∈ Z, hence H ⊂ hσ²i by (8). Conversely, for all w∈F(u, v), q∈Z ⋊ Z and k, l∈Z, we have:

(w;q)(Bk,l; 0,0)(w;q)⁻¹ =(wθ(q)(Bk,l);q)(θ(q⁻¹)(w⁻¹);q⁻¹) = (wθ(q)(Bk,l)w⁻¹; 0,0). (12) Since θ(q)(B_k,l) ∈ Ker(g) and Ker(g) is normal in F(u, v), it follows that wθ(q)(B_k,l)w⁻¹ ∈ Ker(g), and so (w;q)(B_k,l; 0,0)(w;q)⁻¹ ∈ι(Ker(g)) = H by (12). Hence H is a normal subgroup of P₂(K²) by (8), and since σ² ∈ H, we conclude that hσ²i ⊂ H. Thus hσ²i = H as required.

Since ι is injective, the induced homomorphism ι : Ker(g) → hσ²i is an isomorphism, and the image of the basis {Bk,l}k,l∈Z byι yields a basis of the free grouphσ²i. The commutativity of the given diagram follows by considering the images of the elements of this basis{(B_k,l; 0,0)}k,l∈Z and using (10).

Remark 3.4. Although Ker(g) and hσ²i are isomorphic by Proposition 3.3, our results will be stated in terms of hσ²i, since we will formulate most of our equations in this group.

The following result provides a normal form for elements of F(u, v)in terms of g and hσ²i. Proposition 3.5. Let w ∈ F(u, v), and let g(w) = (r, s). Then there exists a unique element x∈ hσ²i such that w=u^rv^sx.

Proof. Letw be as in the statement, and let x=v^−su^−rw. Then w=u^rv^sx, and:

g(x) =g(v^−su^−rw) =g(v)^−sg(u)^−rg(w) = (0,−s)(−r,0)(r, s) = (0,−s)(0, s) = (0,0).

So x∈ hσ²i. Clearly x is unique.

Let p, q ∈Z. Since hσ²i is a normal subgroup ofF(u, v), the following homomorphism is well defined:

c_p,q : hσ²i −→ hσ²i

x 7−→ v^pu^qxu^−qv^−p. (13) Lemma 3.6. Let k, l, p, q ∈ Z, and let (m, n) ∈ Z ⋊ Z. With the notation of Proposition 3.1, there exist γ, λ, η∈ hσ²i such that:

(a) θ(m, n)(B_k,l) = γB_k,ε^εn

nl−2δ_kmγ⁻¹. (b) ρ(Bk,l) = λB₋^εk_k,ε

(k+1)lλ⁻¹. (c) c_p,q(B_k,l) = ηB_k+p,l+εkqη⁻¹.

Proof. During the proof, we will make use freely of Proposition3.1. First, we have:

θ(m, n)(B_k,l) = θ(m, n)(v^ku^lBu^−lv^−k) = θ(m, n)(v^ku^l)B^εnθ(m, n)(v^ku^l)⁻¹ =γB_k,ε^εn

nl−2δ_kmγ⁻¹, where γ = θ(m, n)(v^ku^l)u^εn+1l+2δkmv^−k ∈ F(u, v). To complete the proof of item (a), it remains to show that γ ∈Ker(g). Since:

γ = (B^mvu^−2mB^−m+δn)^k(B^m−δnu^εnB^−m+δn)^lu^εn+1l+2δkmv^−k, and B ∈Ker(g), it follows that:

g(γ) = ((0,1)(−2m,0))^k(ε_nl,0)(ε_n+1l+ 2δ_km,0)(0,−k) = (2m,1)^k(ε_nl+ε_n+1l+ 2δ_km,−k)

= (2δ_km, k)(2δ_km,−k) = (2δ_km+ 2δ_kε_km,0) = (0,0),

using the fact that ε_k = −1 if k is odd, and δ_k = 0 if k is even. Hence γ ∈ Ker(g). To prove item (b), first note that:

θ(g(v^ku^l))(B) = θ((0, k)(l,0))(B) = θ(ε_kl, k)(B) = B^εk. (14) By (11) and (14), we have:

ρ(B_k,l) = ρ(v^ku^lBu^−lv^−k) = ρ(v^ku^l)θ(g(v^ku^l))(B)ρ(v^ku^l)⁻¹ =λB₋^εk_k,ε

(k+1)lλ⁻¹, where λ=ρ(v^ku^l)u^εklv^k∈F(u, v). It remains to show that λ∈Ker(g). We have:

g(λ) = g(ρ(v^ku^l)u^εklv^k) =g((p_F ◦l_σ◦i)(v^ku^l)).(ε_kl, k)

=g(p_F(((uv)^−k(uB)^δk; 0, k)((Bu⁻¹)^lB^−l;l,0))).(ε_kl, k)

=g((uv)^−k(uB)^δkθ(0, k)((Bu⁻¹)^lB^−l)).(ε_kl, k)

= (1,1)^−k(δ_k,0). g((B^εk(B^−δku^εkB^δk)⁻¹)^lB^−εkl).(ε_kl, k)

= (δ_−k,−k)(δ_k,0)(−ε_kl,0)(ε_kl, k) = (δ_−k+ε_−kδ_k,0) = (0,0),

since δ_k = 0 if k is even, and ε_k =−1 if k is odd. Henceλ ∈ Ker(g). Finally we prove item (c).

We have:

c_p,q(B_k,l) =v^pu^qB_k,lu^−qv^−p =v^pu^qv^ku^lBu^−lv^−ku^−qv^−p =ηB_k+p,l+εkqη⁻¹,

where η =v^pu^qv^ku^εk+1qv^−k−p ∈ F(u, v). It remains to show that η ∈Ker(g). This is indeed the case because:

g(η) = (0, p)(q, k)(εk+1q,−k−p) = (0, p)(q+εkεk+1q,−p) = (0, p)(0,−p) = (0,0), which completes the proof.

Let hσ²i_Ab denote the Abelianisation of the grouphσ²i. By abuse of notation, for all k, l∈Z, we denote the image of a generator B_k,l in hσ²iAb byB_k,l. By Proposition 3.3, hσ²iAb is the free Abelian group for which {B_k,l :=v^ku^lBu^−lv^−k |k, l∈Z}is a basis, namely:

hσ²iAb = M

k,l∈Z

Z[B_k,l].

For all (m, n) ∈ Z ⋊ Z and p, q ∈ Z, the endomorphisms θ(m, n), ρ and (c_p,q) of hσ²i induce endomorphisms θ(m, n)Ab, ρAb and (cp,q)Ab of hσ²i_Ab respectively, and by Lemma 3.6, for all k, l ∈Z, they satisfy:

θ(m, n)Ab(B_k,l) =ε_nB_{k,εnl−2δkm} (15)

ρ_Ab(B_k.l) =ε_kB_{−k,ε(k+1)l}, and (16)

(c_p,q)Ab(B_k,l) =B_k+p,l+εkq. (17)

Let k, l ∈ Z and r ∈ {0,1}. If x and y are elements of a group, let [x, y] = xyx⁻¹y⁻¹ denote their commutator. In the rest of the paper, we will be particularly interested in the following elements of F(u, v):

(I) T_k,r =u^k(B^εru^−εr)^kεr. (II) I_k =v^k(vB)^−k.

(III) O_k,l =

v^2k, u^l . (IV) J_k,l =v^2k(vu^l)^−2k.

As we shall now see,Tk,r,Ik,Ok,landJk,l are elements ofhσ²i, and their projections intohσ²i_Abwill be denoted by Te_k,r,Ie_k,Oe_k,l andJe_k,l respectively. The following result describes the decomposition of these Abelianisations in terms of the basis {Bk,l}k,l. Ifl ∈Z, let σl denote its sign,i.e σl= 1 if l > 0,σ_l =−1 if l <0, and σ_l = 0 if l = 0.

Proposition 3.7. Let k, l∈Z and r ∈ {0,1}.

(a) The elements T_k,r, I_k, O_k,l and J_k,l belong to hσ²i.

(b) If k = 0 then Te_0,r =Ie₀ = 0, and if k = 0 or l = 0 then Oe_k,l =Je_k,l = 0.

(c) For all k, l6= 0:

Te_k,r =σ_k

σkk

X

i=1

B_{0,σk(i+(σk(1−2r)−1)/2)}

Iek =−σk σkk

X

i=1

Bσ_ki+(1−σ_k)/2,0

Je_k,l =−σ_kσ_l

σkk

X

i=1 σll

X

j=1

B_{σk(2i−1),σl(j−(1+σl)/2)}

Oe_k,l =σ_kσ_l

σkk

X

i=1 σll

X

j=1

B_{σk(2i−1),−σlj+(σl−1)/2}−B_{σk(2i−1)−1,σlj−(1+σl)/2} .

The proof of Proposition 3.7, which is divided into the following four lemmas, consists essen- tially in manipulating each of the elements to obtain recurrence relations, and using induction to obtain expressions for T_k,r, I_k, O_k,l and J_k,l in hσ²i. Part (c) of Proposition 3.7 is obtained by Abelianising these expressions.

Lemma 3.8. If k ∈Z and r ∈ {0,1} then T_k,r =

σkk

Y

i=1

B_0,k^σk_−σ

ki−r+(σk+1)/2.

Proof. Letr∈ {0,1}. Then clearlyT_0,r =1 and T_1,r =u(B^εru^−εr)^εr =B_0,1−r, so the result holds for k = 1. Suppose that the formula is valid for some k≥1. Then:

T_k+1,r =u^k+1(B^εru^−εr)^(k+1)εr =u. u^k(B^εru^−εr)^kεr.(B^εru^−εr)^εr

=uT_k,ru⁻¹. u(B^εru^−εr)^εr =u Yk i=1

B_{0,k−i+1−r}

!

u⁻¹. B_0,1−r =

k+1Y

i=1

B_{0,k+1−i+1−r},

and so the result holds for all k ≥0 by induction. Suppose that k ≥1. Then:

T_−k,r =u^−k(B^εru^−εr)^−kεr =u^−k(u^k(B^εru^−εr)^kεr)⁻¹u^k=u^−kT_k,r⁻¹u^k

= Yk

i=1

B_0,−i+1−r

!−1

= Yk i=1

B_0,⁻₋¹_k+i−r

using the result for T_k,r in the case k≥1, which proves the formula for all k ∈Z. Lemma 3.9. Let k ∈Z. If k = 0 then I_k =1, and if k 6= 0 then:

I_k=

σ_kk

Y

i=1

B_{i+k(1−σk)/2,0}

!−σk

.

Proof. Ifk = 0, then clearly I₀ =1. If k = 1, I₁ =v(vB)⁻¹ =vB⁻¹v⁻¹ =B_1,0⁻¹, and result holds in this case. Suppose that the formula for I_k holds for some k ≥ 1, and let us prove the formula for k+ 1. We have:

I_k+1 =v^k+1(vB)^−k−1 =vv^k(vB)^−kv⁻¹v(vB)⁻¹ =vI_kv⁻¹I₁

=v Yk i=1

B_i,0

!−1

v⁻¹B_1,0⁻¹ = Yk i=1

B_i+1,0

!−1

B_1,0⁻¹ =

k+1Y

i=1

B_i,0

!−1

,

so by induction, the formula for I_k holds for all k ≥0. If k <0 then −k >0 and so:

I_k=v^k(vB)^−k=v^k(vB)^−kv^kv^−k =v^k v^−k(vB)^k₋1

v^−k =v^kI₋⁻_k¹v^−k

=v^k

−k

Y

i=1

B_i,0

! v^−k =

−k

Y

i=1

B_k+i,0.

It follows that the formula given in the statement holds for all k 6= 0.

Lemma 3.10. Let k, l ∈ Z, let ε∈ {−1,1}, and let ω = (

0 if ε =−1

1 if ε = 1. If k = 0 or l = 0, then J_k,l =1, and if k, l >0, then:

Jk,εl = Yk i=1

Yl j=1

B_2k^−ε_{−2i+1,εj−ω}

!

and J₋k,εl = Yk

i=1

Yl j=1

B₋⁻_2i+1,εj^ε _−ω

!!₋1

.